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find-lim-n-U-n-n-3-2n-2-1-3-n-3-3n-2-1-3-

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Question Number 197479 by pticantor last updated on 19/Sep/23
find: lim_(n→∞) U_n =((n^3 +2n^2 ))^(1/3) −((n^3 −3n^2 ))^(1/3)
$$\boldsymbol{{find}}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{m}}}\:\boldsymbol{{U}}_{\boldsymbol{{n}}} \:=\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{n}}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{\boldsymbol{{n}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{n}}^{\mathrm{2}} }\: \\ $$
Commented by Frix last updated on 19/Sep/23
Strange method but it works u^(1/3) +v^(1/3) =c ⇔ (u−v)^3 −c^3 (3(u^2 +7uv+v^2 )−3c^3 (u−v)+c^6 )=0 (27c^3 −125)n^6 −27c^3 n^5 −87c^3 n^4 −15c^6 n^2 +c^9 =0 ⇒ 27c^3 −125≠0 ⇔ c≠(5/3) lim_(n→∞) U_n =(5/3)
$$\mathrm{Strange}\:\mathrm{method}\:\mathrm{but}\:\mathrm{it}\:\mathrm{works} \\ $$$${u}^{\frac{\mathrm{1}}{\mathrm{3}}} +{v}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c} \\ $$$$\Leftrightarrow \\ $$$$\left({u}−{v}\right)^{\mathrm{3}} −{c}^{\mathrm{3}} \left(\mathrm{3}\left({u}^{\mathrm{2}} +\mathrm{7}{uv}+{v}^{\mathrm{2}} \right)−\mathrm{3}{c}^{\mathrm{3}} \left({u}−{v}\right)+{c}^{\mathrm{6}} \right)=\mathrm{0} \\ $$$$\left(\mathrm{27}{c}^{\mathrm{3}} −\mathrm{125}\right){n}^{\mathrm{6}} −\mathrm{27}{c}^{\mathrm{3}} {n}^{\mathrm{5}} −\mathrm{87}{c}^{\mathrm{3}} {n}^{\mathrm{4}} −\mathrm{15}{c}^{\mathrm{6}} {n}^{\mathrm{2}} +{c}^{\mathrm{9}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{27}{c}^{\mathrm{3}} −\mathrm{125}\neq\mathrm{0}\:\Leftrightarrow\:{c}\neq\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{U}_{{n}} \:=\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Answered by MM42 last updated on 19/Sep/23
lim_( n→∞) (((n+(2/3))^3 +p_2 (n)))^(1/3) −(((n−1)^3 +q_2 (n)))^(1/3) =lim_(n→∞) (n+(2/3))(((1+((p_2 (n))/((n+(2/3))^3 ))))^(1/3) )−(n−1)(((1+((q_2 (n))/((n+(2/3))^3 ))))^(1/3) ) =(2/3)+1=(5/3) ✓
$${lim}_{\:{n}\rightarrow\infty} \:\:\sqrt[{\mathrm{3}}]{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} +{p}_{\mathrm{2}} \left({n}\right)}−\sqrt[{\mathrm{3}}]{\left({n}−\mathrm{1}\right)^{\mathrm{3}} +{q}_{\mathrm{2}} \left({n}\right)} \\ $$$$={lim}_{{n}\rightarrow\infty} \:\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{{p}_{\mathrm{2}} \left({n}\right)}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }}\:\:\right)−\left({n}−\mathrm{1}\right)\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{{q}_{\mathrm{2}} \left({n}\right)}{\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }}\:\:\right)\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\checkmark \\ $$$$ \\ $$
Answered by universe last updated on 19/Sep/23
(1+x)^n = 1+nx+..... where −1<x<1 lim_(n→∞) n[(1+(2/n))^(1/3) − (1−(3/n))^(1/3) ] lim_(n→∞) n[1+(2/(3n))−1+(3/(3n))] lim_(n→∞) n × (5/(3n)) = (5/3)
$$\:\:\:\left(\mathrm{1}+{x}\right)^{{n}} \:=\:\:\mathrm{1}+{nx}+…..\:\:\:\:\:{where}\:−\mathrm{1}<{x}<\mathrm{1} \\ $$$$\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:{n}\left[\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)^{\mathrm{1}/\mathrm{3}} −\:\left(\mathrm{1}−\frac{\mathrm{3}}{{n}}\right)^{\mathrm{1}/\mathrm{3}} \right] \\ $$$$\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:{n}\left[\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}{n}}−\mathrm{1}+\frac{\mathrm{3}}{\mathrm{3}{n}}\right] \\ $$$$\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:{n}\:×\:\frac{\mathrm{5}}{\mathrm{3}{n}}\:\:=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$

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